1. Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. PowerPoint Lecture Presentation by J. David Robertson University of Missouri

2. By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00 amu Atomic mass is the mass of an atom in atomic mass units (amu) Micro World atoms & molecules Macro World grams 3.1

3. Natural lithium is: 7.42% 6 Li (6.015 amu) 92.58% 7 Li (7.016 amu) (7.42 x 6.015) + (92.58 x 7.016) 100 = 6.941 amu 3.1 Average atomic mass of lithium:

4. Average atomic mass (6.941)

5. The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C 3.2 1 mol = N A = 6.0221367 x 10 23 Avogadro’s number (N A )

6. Molar mass is the mass of 1 mole of in grams eggs shoes marbles atoms 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams) 3.2

7. One Mole of: C S Cu Fe Hg 3.2

8. 1 amu = 1.66 x 10 -24 g or 1 g = 6.022 x 10 23 amu 1 12 C atom 12.00 amu x 12.00 g 6.022 x 10 23 12 C atoms = 1.66 x 10 -24 g 1 amu 3.2 M = molar mass in g/mol N A = Avogadro’s number What is the mass, in grams, of one Carbon atom? 1 mole = 12.00 g 1 mole = 6.02 x 10 23 atoms

9. Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 10 23 atoms K 0.551 g K 1 mol K 39.10 g K x x 6.022 x 10 23 atoms K 1 mol K = 8.49 x 10 21 atoms K 3.2

10. Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2 1S32.07 amu 2O+ 2 x 16.00 amu SO 2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2 3.3

11. Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C 3 H 8 O ? 1 mol C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 g C 3 H 8 O 1 mol H = 6.022 x 10 23 atoms H 5.82 x 10 24 atoms H 3.3 1 mol C 3 H 8 O molecules = 8 mol H atoms 72.5 g C 3 H 8 O 1 mol C 3 H 8 O 60 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O x 6.022 x 10 23 H atoms 1 mol H atoms x =

12. KE = 1/2 x m x v 2 v = (2 x KE/m) 1/2 F = q x v x B 3.4 Light Heavy

13. Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C2H6OC2H6O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14%H = 6 x (1.008 g) 46.07 g x 100% = 13.13%O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% 3.5

14. 3.6 g CO 2 mol CO 2 mol Cg C g H 2 O mol H 2 Omol Hg H g of O = g of sample – (g of C + g of H) Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O

15. Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

16. Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N 74.1 g 1 mol 16.00 g = 4.63 mol O 1.85 mol = 1 N = 2.5 O

17. Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5

18. Molecular Formula “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula molecular formula ?

19. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

20. Molecular Formula The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? 28.1 g/mol 14.03 g/mol = 2.00 empirical mass = 14.03 g/mol (CH 2 ) 2  C 2 H 4

21. Find the Molecular Formulas of each of the following compounds A compound with an empirical formula of C 2 OH 4 and a molar mass of 88 grams per mole. A compound with an empirical formula of C 4 H 4 O and a molar mass of 136 grams per mole.

22. 3.7 3 ways of representing the reaction of H 2 with O 2 to form H 2 O A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction reactantsproducts

23. How to “Read” Chemical Equations 2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO 3.7

24. Balancing Chemical Equations 1.Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O 2.Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 3.7 2C 2 H 6 NOT C 4 H 12

25. Balancing Chemical Equations 3.Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O 3.7 start with C or H but not O 2 carbon on left 1 carbon on right multiply CO 2 by 2 C 2 H 6 + O 2 2CO 2 + H 2 O 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3 C 2 H 6 + O 2 2CO 2 + 3H 2 O

26. Balancing Chemical Equations 4.Balance those elements that appear in two or more reactants or products. 3.7 2 oxygen on left 4 oxygen (2x2) C 2 H 6 + O 2 2CO 2 + 3H 2 O + 3 oxygen (3x1) multiply O 2 by 7 2 = 7 oxygen on right C 2 H 6 + O 2 2CO 2 + 3H 2 O 7 2 remove fraction multiply both sides by 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O

27. Balancing Chemical Equations 5.Check to make sure that you have the same number of each type of atom on both sides of the equation. 3.7 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O ReactantsProducts 4 C 12 H 14 O 4 C 12 H 14 O 4 C (2 x 2)4 C12 H (2 x 6)12 H (6 x 2) 14 O (7 x 2)14 O (4 x 2 + 6)

28. 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units Mass Changes in Chemical Reactions 3.8

29. Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x = 235 g H 2 O 3.8

30. 6 green used up 6 red left over Limiting Reagents 3.9

31. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. (this means you will set up 2 stoich problems) 3. Smaller answer indicates: –limiting reactant –amount of product that can be made

32. Limiting Reactants 79.1 g of zinc react with 2.25 mol of HCl. –Identify the limiting and excess reactants. –How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 2.25 mol

33. A. Limiting Reactants 79.1 g Zn 1 mol Zn 65.39 g Zn = 27.1 L H 2 1 mol H 2 1 mol Zn 22.4 L H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 2.25 mol

34. A. Limiting Reactants 22.4 L H 2 1 mol H 2 2.25 mol HCl = 25 L H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 2.25 mol

35. A. Limiting Reactants Zn: 27.1 L H 2 HCl: 25 L H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H 2 left over zinc

36. Finding the Amount of Excess Reactant Left Over 1.calculate the amount of the excess reactant needed to react with the limiting reactant (yet another stoich calculation!) 2.subtract that amount from the given amount to find the amount left over. Zn + 2HCl  ZnCl 2 + H 2 79.1 g 2.25 mol Limiting reactant: HCl Excess reactant: Zn

37. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100 3.10

38. Chemistry In Action: Chemical Fertilizers Plants need: N, P, K, Ca, S, & Mg 3H 2 (g) + N 2 (g) 2NH 3 (g) NH 3 (aq) + HNO 3 (aq) NH 4 NO 3 (aq) 2Ca 5 (PO 4 ) 3 F (s) + 7H 2 SO 4 (aq) 3Ca(H 2 PO 4 ) 2 (aq) + 7CaSO 4 (aq) + 2HF (g) fluorapatite